202211211635 Solution to 1976-CE-AMATH-I-XX

Let the complex number $z_1=-1+\mathrm{i}\sqrt{3}$ .

i. Represent $z_1$ and its conjugate $\overline{z_1}$ on the Argand diagram.
ii. Calculate the modulus and the argument of $z_1$ .
iii. If $z_2=(z_1)^2$ , express $z_2$ in the form of $a+\mathrm{i}b$ .
iv. Show that both $z_1$ and $\displaystyle{\frac{1}{2}z_2}$ are roots of the equation $z^3-8=0$ .

Roughwork.

$\begin{aligned} \textrm{modulus }\mathrm{mod}(z_1) & = |z_1| \\ & = |-1+\mathrm{i}\sqrt{3}| \\ & = \sqrt{(-1)^2+(\sqrt{3})^2} \\ & = 2 \\ \textrm{argument }\mathrm{arg}(z_1) & = \arctan\bigg(\frac{\sqrt{3}}{-1}\bigg) \\ & = \frac{2\pi}{3}\\ \end{aligned}$

Thus, $z_1=2e^{\mathrm{i}\frac{2\pi}{3}}$ .

As is given,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = \big( 2e^{\mathrm{i}\frac{2\pi}{3}}\big)^2 \\ & = 4e^{\mathrm{i}\frac{4\pi}{3}} \\ \mathrm{mod}(z_2) & = 4 \\ \mathrm{arg}(z_2) & = \frac{4\pi}{3} \\ \end{aligned}$

or,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = (-1+\mathrm{i}\sqrt{3})^2 \\ & = (-1)^2 +2(-1)(\mathrm{i}\sqrt{3}) + (\mathrm{i}\sqrt{3})^2 \\ & = 1 -\mathrm{i}(2\sqrt{3}) -3 \\ & = -2-\mathrm{i}(2\sqrt{3}) \\ \end{aligned}$

This problem is not to be attempted.

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